Question

Consider the sum 6+66+66667 where the last addend has 10 digits and the digit in thousands placet square of the digit in hundreds place of the resulting sumi a) 40 b) 19 653 a) 23​

Answer 1

Answer:

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Step-by-step explanation:

Note that the given series is not a geometric series.

We need to find Sn=6+66+666+...to n terms

Sn=6(1+11+111+....to n terms )

=96(9+99+999+....to n terms) (Multiply and divide by 9)

=32[(10−1)+(100−1)+(1000−1)+.... to n terms]

=32[(10+102+103+....n terms)−n]

Thus, Sn=32[910(10n−1)−n]

Answer 2

Step-by-step explanation:

the answer of the question is 23

Note that the given series is not a geometric serial.

We need to find Sn = 6+66+666+...to n terms

Sn = 6 ( 1+11+11+...to n terms )

= 6

__ ( 9+99+999+ ...to n terms)

9

( multiply and divide by 9 )

= 2

__ { ( 10-1 ) + ( 100-1 ) + (1000-1 ) +

3

...to n terms }

= 2

__ { ( 10+10² + 10³+...to n terms ) }

3

thus Sn = 2

_ | 10 (10n - 1 ) |

| _______ - n |

3 | 9 |